Different results for the Hamiltonian of a disc rolling on an inclined plane $\hskip2in$ 
Starting from a Lagrangian of a disc rolling down on a inclined plane without slipping, given by:
$$
\mathcal{L}=\frac{M}{2}\dot{x}^2+\frac{MR^2}{4}\dot{\theta}^2+Mg(x-L)\sin(\alpha) \tag{1}
$$
where $M$ is the disc mass and $\alpha$ is the angle of the inclined plane. Its Hamiltonian in terms of $x$ coordinate momentum $p_x$ and $\theta$ coordinate momentum $p_\theta$ is:
$$
\mathcal{H}=\frac{p_x^2}{2M}+\frac{p_\theta^2}{MR^2}-Mg(x-L)\sin(\alpha) \tag{2}
$$
Since that $\dot{\theta}=\frac{\dot{x}}{R}$, I'm trying to get it's Hamiltonian in terms only on the $x$ coordinate momentum $p_x$, and so far i've got the following two different results:


*

*First Approach

$$
p_\theta=\frac{MR^2}{2}\dot{\theta}=\frac{MR^2}{2}\frac{\dot{x}}{R}
$$
  where
  $$
\dot{x}=\frac{p_x}{M}
$$
  and therefore
  $$
\therefore p_\theta=\frac{R}{2}p_x
$$
  then replacing in the expression for the Hamiltonian (2) it gives:
  $$
\mathcal{H}=\frac{3}{4M}p_x^2-Mg(x-L)\sin(\alpha)
$$


*Second approach

Manipulating the Lagrangian (1) by applying $\dot{\theta}=\frac{\dot{x}}{R}$, it results in
  $$
\mathcal{L}=\frac{3M}{4}\dot{x}^2+Mg(x-L)\sin(\alpha)
$$
  then
  $$
p_x=\frac{\partial\mathcal{L}}{\partial\dot{x}}=\frac{3M}{2}\dot{x}
$$
$$
\dot{x}=\frac{2}{3M}p_x
$$
  Then, the Hamiltonian will be given by
  $$
\mathcal{H}=p_x\cdot\dot{x}-\mathcal{L}
$$
$$
\mathcal{H}=p_x\cdot\frac{2}{3M}p_x-\frac{3M}{4}\Big(\frac{2}{3M}p_x\Big)^2-Mg(x-L)\sin(\alpha)
$$
$$
\mathcal{H}=\frac{2}{3M}p_x^2-\frac{3M}{4}\Big(\frac{4}{9M^2}p_x^2\Big)-Mg(x-L)\sin(\alpha)
$$
$$
\mathcal{H}=\frac{2}{3M}p_x^2-\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)
$$
$$
\mathcal{H}=\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)
$$
  which is a different result that the obtained in the first approach. 

So the question is, which one is the correct, and why the incorrect one is incorrect.
I don't want to say "I guess" but I guess the first one is the correct since the total energy obtained is a little greater than the obtained in the second result, which doesn't makes sense to me that the total energy got reduced
 A: *

*OP is considering a constrained Lagrangian of the form
$$ L(x,\theta;\dot{x},\dot{\theta};\lambda)~=~\frac{M}{2}\dot{x}^2+\frac{I}{2}\dot{\theta}^2 - V(x) -\lambda(x-R\theta). \tag{A}$$ 
The 'reduced' Lagrange equation reads
$$ \left(M+\frac{I}{R^2} \right)\ddot{x}~\approx~-V^{\prime}(x). \tag{B}$$

*OP's second 'reduced' approach is correct:
$$ H_R(x,p_x)~=~\frac{p_x^2}{2(M+I/R^2)}+V(x). \tag{C} $$

*The problem with OP's first approach is that it doesn't properly incorporate the constraint $$x-R\theta~\approx~{\rm const},\tag{D}$$
and its consequence
$$\dot{x}-R\dot{\theta}~\approx~0.\tag{E}$$ 
Eq. (E) means that the Legendre transformation becomes singular if one tries to keep both variables $x$ and $\theta$. This can be done via the Dirac-Bergmann method for constrained systems. The result is
$$ \begin{align} H(x,\theta; p_x, p_{\theta};\lambda,\mu)
&~=~ \frac{p_x^2}{2M} + \frac{p_\theta^2}{2I}+V(x) + \lambda(x-R\theta)+  \mu(p_{\theta}-\frac{I}{MR}p_x)\cr
~\approx~& \frac{p_x^2}{2M} \left(1+\frac{I}{MR^2} \right)+V(x) + \lambda(x-R\theta)+  \mu(p_{\theta}-\frac{I}{MR}p_x). \end{align}\tag{F}$$
The two constraints are of 2nd class. One may check that the corresponding Hamiltonian Lagrangian 
$$ \begin{align}L_H(x,\theta; p_x, p_{\theta};\lambda,\mu)~:=~&p_x\dot{x}+p_{\theta}\dot{\theta}-H(x,\theta; p_x, p_{\theta};\lambda,\mu)\cr 
\quad\stackrel{p_x,p_{\theta},\mu}{\longrightarrow}&\quad L(x,\theta;\dot{x},\dot{\theta};\lambda)\end{align}\tag{G} $$
becomes the original Lagrangian if we integrate out/eliminate the variables $p_x,p_{\theta},\mu$. Or one may check directly that Hamilton's equations becomes Lagrange's equations if we eliminate the variables $p_x,p_{\theta},\mu$.
